Let us consider the smooth map \(f\times f: M\times M\rightarrow \mathbb {R}^2\)  defined as \((f\times f)(x,y)=(f(x),f(y))\) . As \(f^{-1}[a,b]\)  is compact we get that \((f\times f)^{-1}[a,b]\times [a,b]=f^{-1}[a,b]\times f^{-1}[a,b]\)  is compact in \(M\times M\)  and because of this \(\rho ^{-1}(f^{-1}[a,b]\times f^{-1}[a,b])\)  is compact in \(G\)  since \(\rho \)  is a proper map. Another straightforward computation allows us to conclude that \(\rho ^{-1}(f^{-1}[a,b]\times f^{-1}[a,b])=(s^\ast f)^{-1}[a,b]\)  since \(s^\ast f=t^\ast f\)  and therefore \((s^\ast f)^{-1}[a,b]\)  is compact in \(G\) .
Let us now follow some ideas stated in [1]}, [2]} about the proof of this result in the classical case. It is clear that if \(f^{-1}[a,b]\cap \textnormal {Crit}(f)=\emptyset \) , then \((s^\ast f)^{-1}[a,b]\cap \textnormal {Crit}(F)=\emptyset \) . We know that \(\widetilde{X}=\nabla (f\circ s)\)  and \(X=\nabla f\)  are gradient-like vector fields on \(G\)  and \(M\) , respectively. Thus, consider the smooth function \(\mu :M\rightarrow [0,\infty )\)  which is defined by \(\Vert  Xf\Vert ^{-1}\)  in \(f^{-1}[a,b]\)  and that vanishes outside a compact neighborhood of this set. We can similarly construct a smooth function \(\widetilde{\mu }:G\rightarrow [0,\infty )\)  by using instead \(\widetilde{X}\)  and \((s^\ast f)^{-1}[a,b]\) . On the one hand, observe that since \(\nabla F=(\widetilde{X},X)\)  is a multiplicative vector field we get that
\(\mu (s(g))& = & \Vert  (Xf)(s(g))\Vert ^{-1}=\Vert  df(s(g))(X(s(g))\Vert ^{-1}\\& = & \Vert  df(s(g))(ds(g)(\widetilde{X}(g)))\Vert ^{-1}=\Vert  d(f\circ s)(g)(\widetilde{X}(g))\Vert ^{-1}\\& = & \Vert  (\widetilde{X}F)(g)\Vert ^{-1}=\widetilde{\mu }(g),\) 
what means that we have \(\widetilde{\mu }=\mu \circ s\) . On the other hand, let us consider the vectors fields \(-\widetilde{\mu }\widetilde{X}\)  and \(-\mu X\) . From [1]} we obtain that they are complete vector fields. Furthermore, the pair \((-\widetilde{\mu }\widetilde{X},-\mu X)\)  defines again a multiplicative vector field. Indeed, as consequence of Proposition REF  we get
\((ds\circ \widetilde{\mu }\widetilde{X})(g)=\widetilde{\mu }(g)ds(g)(\widetilde{X}(g))=\widetilde{\mu }(g)X(s(g))=\mu (s(g))X(s(g))=(\mu X\circ s)(g).\) 
Given that \(\nabla (f\circ s)=\nabla (f\circ t)\)  we actually have that \(\widetilde{\mu }=\mu \circ s=\mu \circ t\)  and thus we can analogously obtain that \(dt\circ \widetilde{\mu }\widetilde{X}=\mu X\circ t\) . To prove the identity regarding the composition map observe that the formulas \(s\circ m=s\circ \pi _2\)  and \(t\circ m=t\circ \pi _1\)  imply that \(\widetilde{\mu }(m(g,h))=\widetilde{\mu }(g)=\widetilde{\mu }(h)\)  for all \((g,h)\in G^{(2)}\) . Therefore,
\((dm\circ (\widetilde{\mu }\widetilde{X}\times \widetilde{\mu }\widetilde{X}))(g,h) &=& dm_{(g,h)}(\widetilde{\mu }(g)\widetilde{X}(g), \widetilde{\mu }(h)\widetilde{X}(h))\\&=& \widetilde{\mu }(m(g,h))dm_{(g,h)}(\widetilde{X}(g),\widetilde{X}(h))\\&=& \widetilde{\mu }(m(g,h))\widetilde{X}(m(g,h))\\&=& (\widetilde{\mu }\widetilde{X} \circ m)(g,h).\) 
For all \(\tau \in \mathbb {R}\) , let \(\widetilde{\Phi }_{\tau }:G\rightarrow G\)  and \(\Phi _{\tau }:M\rightarrow M\)  denote the respective flows generated by the pair \((-\widetilde{\mu }\widetilde{X},-\mu X)\) . From [4]} we know that \((\widetilde{\Phi }_{\tau },\Phi _{\tau })\)  is a Lie groupoid isomorphism for all \(\tau \in \mathbb {R}\) . As in the classical case (see [1]}, [2]}) we get diffeomorphisms
\(\Phi _{b-a}(M^b)=M^a,\qquad \Phi _{a-b}(M^a)=M^b\quad \textnormal {and}\quad \widetilde{\Phi }_{b-a}(G^b)=G^a,\qquad \widetilde{\Phi }_{a-b}(G^a)=G^b\) 
so that \((\widetilde{\Phi }_{b-a},\Phi _{b-a}):(G^b\rightrightarrows M^b)\rightarrow (G^a\rightrightarrows M^a)\)  defines a Lie groupoid isomorphism with obvious inverse \((\widetilde{\Phi }_{a-b},\Phi _{a-b})\) .
Finally, we can define two deformation retractions \(H:[0,1]\times M^b\rightarrow M^b\)  and \(\widetilde{H}:[0,1]\times G^b\rightarrow G^b\)  respectively as
\(H(\tau ,x)=\Phi _{\tau \cdot (f(x)-a)^+}(x)\qquad \textnormal {and}\qquad \widetilde{H}(\tau ,g)=\widetilde{\Phi }_{\tau \cdot ((s^\ast f)(g)-a)^+}(g).\) 
Here \(r^+:=\max \lbrace r,0\rbrace \)  for every real number \(r\) . Hence, we obtain again a Lie groupoid morphism which allows us to conclude that \(G^a\rightrightarrows M^a\)  is a deformation retraction of \(G^b\rightrightarrows M^b\) .
